Inversion theorem for Dirichlet series Can someone come up with a proof for this little theorem?
Suppose that $F_a(s)$ is a Dirichlet series and $a(n)$ is its associated arithmetic function, that is:
$$F_a(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^s}$$
Then the $a(n)$ are given by:
\begin{equation} \label{eq:a(n)} \nonumber
a(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!}
\end{equation}
The advantage of this formula is that if you know $F_a(s)$ at the even integers, you know the coefficients of its series expansion.

Let me give an example, before questions rain down asking for clarification. 
We know that:
$$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$
where $\mu(n)$ is the Mobius function. Therefore:
$$\mu(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-1}}{(2i+1-2j)!}$$
Now, one good application of this formula is the generalization of the Mobius function. Whatever the $k$, real or complex, positive or negative, we have by definition that:
$$\zeta(s)^{-k}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$$
and you may ask, what is the $\mu_k(n)$ for each $n$ that satisfies the above equation?
Well, they are given by:
$$\mu_k(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-k}}{(2i+1-2j)!}$$
An example:
$\sqrt{\zeta(s)}=1+\frac{1}{2\cdot 2^s}+\frac{1}{2\cdot 3^s}+\frac{3}{8\cdot 4^s}+\frac{1}{2\cdot 5^s}+\frac{1}{4\cdot 6^s}+\frac{1}{2\cdot 7^s} +\frac{5}{16\cdot 8^s}
+\frac{3}{8\cdot 9^s}+\frac{1}{4\cdot 10^s}+\frac{1}{2\cdot 11^s}+\cdots\\$
Another example, the Von Mangoldt function divided by the log is given by:
$$\frac{\Lambda(n)}{\log n}=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\log\zeta(2j)}{(2i+1-2j)!}\\$$
 A: First let's prove this for $$a(n)=\delta_{m,n}=\begin{cases}1,&n=m\\0,&n\neq m\end{cases}$$
with a (fixed) positive integer $m$. Consider the RHS of the formula,
$$A_{m,n}=-2\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{-2j}}{(2i+1-2j)!}.$$
Replacing $j$ by $i-j$ in the inner sum, we get
$$A_{m,n}=-2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}.\tag{1}\label{eq1}$$
Since $\displaystyle\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=\frac{\sin 2\pi m}{2\pi m}=0$, this implies
$$A_{m,n}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i+1}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}$$
which, after replacing $j$ with $i+j$, gives
$$A_{m,n}=2\sum_{i,j=0}^{\infty}\frac{(-1)^{i+j}(2\pi)^{2i+2j}n^{2i}m^{2j}}{(2i+2j+1)!}.\tag{2}\label{eq2}$$
In particular, $A_{m,n}=A_{n,m}$. Now if $n<m$ then we can change the order of summation in $\eqref{eq1}$:
$$A_{m,n}=-2\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}\sum_{i=j}^{\infty}\Big(\frac{n}{m}\Big)^{2i}=-2\Big(1-\frac{n^2}{m^2}\Big)^{-1}\frac{\sin 2\pi n}{2\pi n}=0,$$
and if $n=m$ then, collecting terms in $\eqref{eq2}$ with $i+j=k$, we obtain
$$A_{n,n}=2\sum_{k=0}^{\infty}\frac{(-1)^k(2\pi n)^{2k}}{(2k+1)!}(k+1)=\cos 2\pi n+\frac{\sin 2\pi n}{2\pi n}=1.$$
Concluding, we have $A_{m,n}=\delta_{m,n}$, and this finishes the proof for $a(n)=\delta_{m,n}$.

Essentially, we've proven the formula for functions $a(n)$ such that $\{n:a(n)\neq 0\}$ is finite. It's also easy to see that the formula holds when $\sum_{n=1}^{\infty}a(n)$ converges absolutely: the triple sum
$$\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi)^{-2j}}{(2i+1-2j)!}\sum_{\color{blue}{m=n+1}}^{\infty}\frac{a(m)}{m^{2j}}$$
converges absolutely in this case, so we can change the order of summation, but each of the sums with a fixed value of $m$ equals $0$ as already proven; so the entire sum is $0$, and we're left with $\sum_{m=1}^{n}$, which again has been treated already.
Further results require analytic continuation. This can be done as usual (beginning with application of the above to $\bar{a}(n)=n^{-s}a(n)$ with $\Re s>\sigma$ the abscissa of absolute convergence of $F_a(s)$), but behaviour of the series in the RHS of the formula needs to be explored separately.
A: This proof is simple:
\begin{equation} \nonumber
-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \Rightarrow
\end{equation}
\begin{equation} \nonumber
-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}}{(2i+1-2j)!}\sum_{k=1}^{\infty}\frac{a(k)}{k^{2j}} \Rightarrow
\end{equation}
\begin{equation} \nonumber
\sum_{k=1}^{\infty}a(k)\left(-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}k^{-2j}}{(2i+1-2j)!}\right)\\
\end{equation}
The theorem then follows from the following equation:
\begin{equation} \nonumber
-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi k)^{-2j}}{(2i+1-2j)!}=\begin{cases}
      1, & \text{if}\ n=k\\
      0, & \text{otherwise}\\
\end{cases} 
\end{equation}
And the above equation is justified for being the convolution of $\mu_0(n)$ (which is the same as $\delta_{n1}$) and the associated function of the series $k^{-s}$, $b(n)$, since from the convolution formula:
\begin{equation} \nonumber
c(n)=(\mu_0*b)(n)=\sum_{d|n}\mu_0(d) b\left(\frac{n}{d}\right)=b(n)=\begin{cases}
      1, & \text{if}\ n=k\\
      0, & \text{otherwise}
\end{cases} 
\end{equation}
